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I'm working on this problem:

Let: $$ f(x,y) = \begin{cases} \frac{x^{3/5}y^{2/5}}{\sqrt{x^2+y^2}} &\text{if} \ (x,y) \neq (0,0)\\ 0 &\text{if} \ (x,y) = (0,0) \end{cases} $$

I want to show that $f$ is continuous at the origin.

So my scratch/set up is:

$\forall\varepsilon>0, \exists\delta>0 \ s.t. \ |(x,y)-(\overline{x},\overline{y})|<\delta \implies |f(x,y) - f(\overline{x},\overline{y})|<\varepsilon$

So I get:

$|f(x,y) - f(\overline{x},\overline{y})|=\left|{\cfrac{x^{3/5}y^{2/5}}{\sqrt{x^2+y^2}}-0}\right|<\varepsilon$

If we convert to polar coordinates and utilise the Pythagorean Identity:

$x = r\cos{\theta}, \ \ \ y = r\sin{\theta}$

The function should reduce down to (an image of the steps taken are below):

$\left|{\cfrac{x^{3/5}y^{2/5}}{\sqrt{x^2+y^2}}-0}\right|=(\cos^{3}{\theta}\cdot\sin^{2}{\theta})^{1/5}<\varepsilon$

If we were working in the (x,y) plane the sinusoidal functions would be bounded above by 1, however I'm not 100% sure what they are bounded by in polar form. Am I on the right track for showing continutiy?

SCRATCH WORKINGS:

enter image description here

Sprawk48
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1 Answers1

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You arrive at the conclusion that for any $(x, y) = (r\cos\theta, r\sin\theta)\neq (0, 0)$, you have that $$ f(x, y) = (\cos^3\theta\cdot \sin^2\theta)^{1/5} $$ for arbitrary $\theta$. What happens if I choose $\theta = \pi/4$? What about $\theta = 0$? What does this tell you about the continuity at the point $(0, 0)$?

  • My gut is telling me that at $(0,0)$ the continuity at this point isn't dependant on $\theta$, I think I am getting the idea in my head but finding it hard to put to words/logic – Sprawk48 Aug 31 '22 at 02:36
  • If I look at the sequence $(x_n, y_n) = (1/n \cos(0), 1/n \sin(0)) = (1/n, 0)$, then note that $(x_n, y_n)\to (0, 0)$ yet all $(x_n, y_n)\neq (0, 0)$. By the sequential characterization of continuity, we expect to see that $\lim_{n\to\infty}f(x_n, y_n) = f(0, 0) = 0$, and here that is indeed what we see. However, we expect to see this for any old sequence $(x_n, y_n)$ converging to zero. What happens if we instead look at the sequence $(x_n', y_n') = (1/n \cos(\pi/4), 1/n \sin(\pi/4)) = \left(\frac{1}{n\sqrt{2}} , \frac{1}{n\sqrt{2}} \right)$? What happens to $\lim_{n\to\infty}f(x_n', y_n')$? –  Aug 31 '22 at 02:45
  • Oh! It still converges to $(0,0)$, and that would be because sin and cosine are bounded by 1, so regardless of what value we have, the convergence will eventually behave like $1/n$ for sufficiently large n. – Sprawk48 Aug 31 '22 at 02:52
  • @Sprawk48 ???? Please repeat your calculations for the above case. $x$ and $y$ go to zero, but the function does not. – Andrei Aug 31 '22 at 03:14
  • @Andrei Right, I got ahead of myself. Yeah, $\lim_{n\to \infty} f(x'_n,y'_n) \neq 0$ from the second case. I still feel like I'm at a loss for how to bring this logic to show that the original function is continuous. – Sprawk48 Aug 31 '22 at 03:59
  • @Sprawk48 the point is that when $x=y$, no matter how small is $x>0$, the function is equal to $1/\sqrt(2)\approx 0.707$. Then, I can choose $\varepsilon=0.1$. There is no $\delta$ such that if $|(x,x)|<\delta$ then $|f(x,x)-0|<\varepsilon$. So the function has a discontinuity. – Andrei Aug 31 '22 at 12:32